Match List I with List II

LIST I		LIST II
A.	Slope of the tangent to curve x3 − 2x at x = 2	I.	−81
B.	Slope of line passing through the points (0, 2) and (5, −6)	II.	10
C.	Point at which the tangent to the curve y = √4x − 3 has its slope = 2/3	III.	−8/5
D.	Slope of normal to the curve y = x−2 x−1 at x = 10	IV.	(3, 3)

Choose the correct answer from the options given below:

The correct option is A-II, B-III, C-IV, D-I.

Let us verify each match step-by-step:

Step 1: Match for A
We need to find the slope of the tangent to the curve $y=x^3-2x$ at the point where $x=2$.
The slope of the tangent to a curve $y=f\left(x\right)$ at any point is given by the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$.
Differentiating $y=x^3-2x$ with respect to $x$:
$\frac{dy}{dx}=3x^2-2$
Now, we substitute $x=2$ into the derivative to find the slope at that point:
${\left[\frac{dy}{dx}\right]}_{x=2}=3{\left(2\right)}^2-2=3\left(4\right)-2=12-2=10$
Thus, A matches with II.

Step 2: Match for B
We need to find the slope of the line passing through the points $(0,2)$ and $(5,-6)$.
The formula for the slope $m$ of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is:
$m=\frac{y_2-y_1}{x_2-x_1}$
Substituting the given coordinates $(x_1,y_1)=(0,2)$ and $(x_2,y_2)=(5,-6)$ into the formula:
$m=\frac{-6-2}{5-0}=\frac{-8}{5}=-\frac{8}{5}$
Thus, B matches with III.

Step 3: Match for C
We need to find the point on the curve $y=\sqrt{4x-3}$ where the slope of the tangent is $\frac{2}{3}$.
Let us find the derivative of the curve with respect to $x$:
$\frac{dy}{dx}=\frac{d}{dx}{(4x-3)}^{\frac{1}{2}}=\frac{1}{2}{(4x-3)}^{-\frac{1}{2}}\cdot 4=\frac{2}{\sqrt{4x-3}}$
We set this slope equal to $\frac{2}{3}$:
$\frac{2}{\sqrt{4x-3}}=\frac{2}{3}$
Dividing both sides by 2 gives:
$\frac{1}{\sqrt{4x-3}}=\frac{1}{3}\Rightarrow \sqrt{4x-3}=3$
Squaring both sides:
$4x-3=9\Rightarrow 4x=12\Rightarrow x=3$
Now, find the corresponding y-coordinate on the curve:
$y=\sqrt{4\left(3\right)-3}=\sqrt{12-3}=\sqrt{9}=3$
So, the point is $(3,3)$.
Thus, C matches with IV.

Step 4: Match for D
We need to find the slope of the normal to the curve $y=\frac{x-2}{x-1}$ at $x=10$.
First, find the derivative $\frac{dy}{dx}$ using the quotient rule:
$\frac{dy}{dx}=\frac{(x-1)\frac{d}{dx}(x-2)-(x-2)\frac{d}{dx}(x-1)}{{(x-1)}^2}$
$\frac{dy}{dx}=\frac{(x-1)\left(1\right)-(x-2)\left(1\right)}{{(x-1)}^2}=\frac{x-1-x+2}{{(x-1)}^2}=\frac{1}{{(x-1)}^2}$
At $x=10$, the slope of the tangent ($m_t$) is:
$m_t=\frac{1}{{(10-1)}^2}=\frac{1}{9^2}=\frac{1}{81}$
The slope of the normal ($m_n$) is the negative reciprocal of the slope of the tangent:
$m_n=-\frac{1}{m_t}=-81$
Thus, D matches with I.